Add and Simplify

[NoGoodAtMath]

Does anyone know how to solve this? I think the denominators are (y+1)(y+1)(y-1), so do I add the (y-1) to the top to make a 4y^2+3?

(2y/y^2+2x+1) +(y/y^2-1)

 

[Deleted member 4993]

Does anyone know how to solve this? I think the denominators are (y+1)(y+1)(y-1), so do I add the (y-1) to the top to make a 4y^2+3?

(2y/y^2+2x+1) +(y/y^2-1)

Correct - then follow the method shown to you at:

http://www.freemathhelp.com/forum/threads/80923-Subtract-and-Simplify?p=333864#post333864

 

[lookagain]

You're still not careful with brackets; should be:
2y / (y^2+ > > 2x < < +1) + y / (y^2-1)

No, it should be 2y/(y^2 + 2y + 1) + y/(y^2 - 1).

I think the denominators are (y+1)(y+1)(y-1),

The denominators are equivalent to (y + 1)(y + 1) and (y + 1)(y - 1), respectively.


so do I add the (y-1) to the top to make a

4y^2+3? Where/how did you get the "4y^2 + 3?"

\(\displaystyle \dfrac{2y}{y^2 + 2y + 1} \ + \ \dfrac{y}{y^2 - 1} \ = \)


\(\displaystyle \dfrac{2y}{(y + 1)(y + 1)} \ + \ \dfrac{y}{(y + 1)(y - 1)} \ = \)


\(\displaystyle \dfrac{2y(y - 1)}{(y + 1)(y + 1)(y - 1)} \ + \ \dfrac{y(y + 1)}{(y + 1)(y - 1)(y + 1)} \ = \)


\(\displaystyle \dfrac{2y^2 - 2y}{(y + 1)(y + 1)(y - 1)} \ + \ \dfrac{y^2 + y}{(y + 1)(y - 1)(y + 1)} \ = \)


\(\displaystyle \dfrac{2y^2 - 2y + y^2 + y}{(y + 1)(y + 1)(y - 1)} \)


NoGoodAtMath,

now please continue from that.